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October 14th, 2015, 09:57 PM   #1
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Projection Transformation on x Axis Parallel to y=2x

Hello,

Let $T \colon \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the projection transformation on x axis in parallel to $y=2x$. Find the representing matrix of $T^*$ for the standard basis $B=\left(\left \{1,0 \right \},\left\{0,1\right\}\right)$.

I have a solution which I don't understand.

A basis for $y=2x$ could be $\left \{(1,2) \right\}$. A basis for x axis could be $\left \{(1,0)\right\}$. Therefore A basis for the whole vector would be $$C=\left\{(1,0)\right\},\left\{(1,2)\right\}$$.

I understand why this basis is correct. However:

$e_1=(1,0)=1v_1+0v_2 \Rightarrow [e_1]_B=\left (\begin {matrix} 1\\ 0 \end {matrix}\right)$ - this is quite clear.

$e_2=(0,1)=-\frac{1}{2}v_1+\frac{1}{2}v_2\Rightarrow [e_2]_B=\left (\begin {matrix} -\frac{1}{2} \\ \frac{1}{2} \end {matrix}\right)$ - this is the part I don't understand. Why use $-\frac{1}{2}$ to multiply $v_1$? What am I missing here?

Next, since the proejction transformation is $T(x,y)=(x,0)$ it is easy enough to continue. But there's that part I don't understand. Why is the scalar before the vector $v_1$has to be$
-\frac{1} {2} $

Thanks,

Alan
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October 15th, 2015, 09:20 AM   #2
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Given unit vector u. Projection of any vector x onto u in matrix form is given by

Tx where $\displaystyle T_{ij} =u_{i}u_{j}$

Change of basis:
e1'=P11e1+P21e2
e2'=P12e1+P22e2

Relative to new basis:
$\displaystyle T'=P^{-1}TP$

Reference:
https://www.khanacademy.org/math/lin...ix-vector-prod

Linear Transformations.

EDIT:
I think I see now what you are asking: Find the projection of x onto x axis in direction of line y=2x. Have to think about how to put that in matrix form.

Last edited by zylo; October 15th, 2015 at 09:28 AM.
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October 15th, 2015, 10:08 AM   #3
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Projection of x in direction u on x axis.

ae1+bu=x and solve for a
a=x1-x2u1/u2

In matrix form:
Tx=
$\displaystyle \begin{bmatrix}
1 &-u1/u2 \\
0& 0
\end{bmatrix}\begin{bmatrix}
x1\\
x2
\end{bmatrix}=\begin{bmatrix}
x1-x2u1/u2\\
0
\end{bmatrix}$

Last edited by zylo; October 15th, 2015 at 10:20 AM.
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October 19th, 2015, 09:47 AM   #4
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Thank you both!
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October 20th, 2015, 08:33 AM   #5
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Quote:
Originally Posted by Nerazzurri10 View Post
Thank you!
You're welcome.
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October 31st, 2015, 07:17 AM   #6
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Systematic Solution to OP based on first principles.

T:Rn->Rm is a linear transformation if T(au+bv)=aTu+bTv

Matrix equivalent of T:R2->R2
T(x1e1+x2e2)=x1Te1+x2Te2=x1(T11e1+T21e2)+x2(T12e1+ T22e2)
Tij are the matrix components of T
Tx=[T]x

Example: Project x onto line w in direction v
u: unit vector in direction of x
v: unit vector in direction of projection
w: unit vector along line which x is to be projected onto,
au+bv=cw (draw a sketch, all vectors from origin)

Te1=cw when au=e1
e1+bv=cw
1+bv1=cw1
bv2=cw2, and solve:
c=v2/(w1v2-w2v1)

Te2=c'w when a'u=e2
e2+b'v=c'w
b'v1=c'e1
1+b'v2=c'w2, and solve
c'=-v1/(w1v2-w2v1)

Te1=cw=c(w1e1+w2e2)=T11e1+T21e2
Te2=c'w=c'(w1e1+w2e2)=T12e1+T22e2

T11=cw1, T21=cw2, T12=c'w1, T22=c'w2

EDIT: That's the general case. For OP line to be projected onto is x axis so w1=1, w2=0, and also v1=1/sqrt(5), v2=2/sqrt(5)

Last edited by zylo; October 31st, 2015 at 07:30 AM.
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